student conducted an experiment and find the density of an ICEBERGE. A students than recorded the following readings. Mass 425 25   g Volume 405 15   mL What experimental value should be quoted for the density of the ICEBERG? Compare your answer with the density of water, which is 3 1.00 10  kg . Show any calculations necessary to justify your answer

Answer:

6 gallons

Explanation:

At 30 mph, the fuel mileage is 25 mpg.

After 5 hours, the distance traveled is:

30 mi/hr × 5 hr = 150 mi

The amount of gas used is:

150 mi × (1 gal / 25 mi) = 6 gal

Answer:

Explanation:

  a )

from lens makers formula

[tex]\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})[/tex]

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

[tex]\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})[/tex]

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens  f  = 1.8 cm

image distance  v   = ?

lens formula

[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]

[tex]\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}[/tex]

[tex]\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}[/tex]

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

Answer:

0.938 seconds

Explanation:

For the ball thrown upwards, we use the formula below to solve it:

[tex]s = ut - \frac{1}{2}gt^2[/tex]

where s = distance moved

u = initial speed = 19.2 m/s

t = time taken

g = acceleration due to gravity = 9.8 [tex]m/s^2[/tex]

Let x be the height at which both balls are level, this means that:

=> [tex]x = 19.2t - 4.9t^2[/tex]________(1)

For the ball dropped downwards, we use the formula below:

[tex]s = ut + \frac{1}{2}gt^2[/tex]

u = 0 m/s

At the point where both balls are level:

s = 18 - x

=> [tex]18 - x = 0 + 4.9t^2[/tex]

=> [tex]x = 18 - 4.9t^2[/tex]__________(2)

Equating both (1) and (2):

[tex]19.2t - 4.9t^2 = 18 - 4.9t^2\\\\=> 19.2t = 18\\\\t = 18/19.2 = 0.938 secs[/tex]

They will be level after 0.938 seconds

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration  of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

The [tex]W_x[/tex] is a x-component of force due to gravity (W) and, in this case, is given by: [tex]W_x[/tex] = W.sin(14)

W is described as: W = m.g

Force due to friction ([tex]f_s[/tex]) is given by: [tex]f_s[/tex] = μs.N

N is the normal force and, in the system, is equivalent of [tex]W_y[/tex], so:

[tex]W_y[/tex] = m.g.cos(14)

Therefore, the formula will be:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

Answer:

a. the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal

b. 0.847 m

Explanation:

a. Using v² = u² + 2as, we find the initial vertical velocity of the cat. Now at the peak height, v = final velocity = 0, u = initial velocity and a = -g = 9.8 m/s², s vertical distance travelled by the cat from its position on the cabinet = Δy = 3.130 m - 1.765 m = 1.365 m.

Substituting these variables into the equation, we have

0² = u² + 2(-9.8m/s²) × 1.365 m

-u² = -26.754 m²/s²

u = √26.754 m²/s²

u = 5.17 m/s

To find its initial horizontal velocity, u₁ we first find the time t it takes to reach the peak height from

v = u + at. where the variables mean the same as above.

substituting the values, we have

0 = 5.17 m/s +(-9.8m/s²)t

-5.17 m/s = -9.8m/s²t

t = -5.17 m/s ÷ (-9.8m/s²)

= 0.53 s

Now, the horizontal distance d = u₁t = 1.560 m

u₁ = d/t = 1.560 m/0.53 s = 2.96 m/s

So, the initial velocity of the cat is V = √(u² + u₁²)

= √((5.17 m/s)² + (2.96 m/s)²)

= √(26.729(m/s)² + 8.762(m/s)²)

= √(35.491 (m/s)²)

= 5.95 m/s

its direction θ = tan⁻¹(5.17 m/s ÷ 2.96 m/s) = 60.2°

So, the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal

(b)

First, we find the time t' it takes the cat to land on the bed from d' = u₁t'

where d' = horizontal distance of cabinet from bed = 3.582 m

u₁ = horizontal velocity = 2.96 m/s

t' = d'/u₁

= 3.582 m/2.96 m/s

= 1.21 s

The vertical between the bed and cabinet which is the vertical distance moved by the cat is gotten from Δy = ut' +1/2at'²

substituting u = initial vertical velocity = 5.17 m/s, t' = 1.21 s and a = -g = -9.8 m/s² into Δy, we have  

Δy = ut' +1/2at'² = 5.17 m/s × 1.21 s +1/2(- 9.8 m/s²) × (1.21 s)² = 6.256 - 7.174 = -0.918 m

Δy = y₂ - y₁

Since our initial position is the position of the cabinet above the ground = y₁ = 1.765 m

y₂ = position of bed above ground.

Δy = y₂ - y₁ = -0.918 m

y₂ - 1.765 m = -0.918 m

y₂ = 1.765 m - 0.918 m

= 0.847 m

t\12 and the parties are spreading ever

Explanation:

my point is that you can get sick if

you sont wash your ha

nds or be

save

Answer:

a) a = 5.03x10¹³ m/s²

b) [tex]V_{f} = 4.4 \cdot 10^{5} m/s [/tex]

Explanation:    

a) The acceleration of the positron can be found as follows:

[tex] F = q*E [/tex]    (1)

Also,

[tex] F = ma [/tex]    (2)

By entering equation (1) into (2), we have:

[tex] a = \frac{F}{m} = \frac{qE}{m} [/tex]

Where:

F: is the electric force

m: is the particle's mass = 9.1x10⁻³¹ kg

q: is the charge of the positron = 1.6x10⁻¹⁹ C    

E: is the electric field = 286 N/C

[tex] a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2} [/tex]

b) The positron's speed can be calculated using the following equation:

[tex] V_{f} = V_{0} + at [/tex]

Where:

[tex]V_{f}[/tex]: is the final speed =?

[tex]V_{0}[/tex]: is the initial speed =0

t: is the time = 8.70x10⁻⁹ s

[tex] V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s [/tex]

I hope it helps you!

Answer:2000

Explanation:

Answer:

Explanation:

1 ) Since it is a isochoric process , heat energy passed into gas

= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .

= 7.4 x 12.47 x ( 500 - 300 )

= 18455.6 J.

2 ) Since there is no change in volume , work done by the gas is constant.

3 ) from  , gas law equation

PV = nRT

P = nRT / V

= 7.4 x 8.3 x 500 / .74

= .415 x 10⁵ Pa.

4 ) Average kinetic energy  of gas molecules after attainment of final temperature

= 3/2 x R/ N x T

= 1.5 x 1.38  x 10⁻²³ x 500

= 1.035 x 10⁻²⁰ J

1/2 m v² = 1.035 x 10⁻²⁰

v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶

= 1.49 x 10⁶

v = 1.22 x 10³ m /s

5 )  In this process , pressure remains constant

gas is cooled from 500 to 300 K

heat will be withdrawn .

heat withdrawn

= n Cp dT

= 7.4 x 20.79 x 200

= 30769.2 J .

6 )

gas will have reduced volume due to cooling

reduced volume = .74 x 300 / 500

= .444 m³

change in volume

= .74 - .444

= .296 m³

work done on the gas

= P x dV

pressure x change in volume

= .415 x 10⁵ x .296

= 12284 J.

Answer:

B. energy

Explanation:

A vector has direction.

Energy does not have a direction.

Answer:weight

Explanation:weight

Answer:

a) 98 m/s

b) 19.6 s

c)  1449.8 m

d)  29.6 s

e)  168.6 m/s

f)  46.8 s

Explanation:

Given that

Acceleration of the rocket, a = 5 m/s²

Altitude of the rocket, s = 960 m

a)

Using the equation of motion

v² = u² + 2as, considering that the initial velocity, u is 0. Then

v² = 2as

v = √2as

v = √(2 * 5 * 960)

v = √9600

v = 98 m/s

b)

Using the equation of motion

S = ut + ½at², considering that initial velocity, u = 0. So that

S = ½at²

t² = 2s/a

t² = (2 * 960) / 5

t² = 1920 / 5

t² = 384

t = √384 = 19.6 s

c)

Using the equation of motion

v² = u² + 2as, where u = 98 m/s, a = -9.8 m/s², so that

0 = 98² + 2(-9.8) * s

9600 = 19.6s

s = 9600/19.6

s = 489.8 m

The maximum altitude now is

960 m + 489.8 m = 1449.8 m

d)

Using the equation of motion

v = u + at, where initial velocity, u = 98 m, a = -9.8 m/s. So that

0 = 98 +(-9.8 * t)

98 = 9.8t

t = 98/9.8

t = 10 s

Total time then is, 10 + 19.6 = 29.6 s

e) using the equation of motion

v² = u² + 2as, where initial velocity, u = o, acceleration a = 9.8 m/s, and s = 1449.8 m. So that,

v² = 0 + 2 * 9.8 * 1449.8

v² = 28416.08

v = √28416.08

v = 168.6 m/s

f) using the equation of motion

S = ut + ½at², where s = 1449.8 m and a = 9.8 m/s

1449.8 = 0 + ½ * 9.8 * t²

2899.6 = 9.8t²

t² = 2899.6/9.8

t² = 295.88

t = √295.88

t = 17.2 s

total time in air then is, 17.2 + 29.6 = 46.8 s

Answer:

a) Ff = 19.29 N

b) v = 3.00 m/s

Explanation:

a) To calculate the friction force you use the second Newton Law in the incline plane, with an acceleration equal to zero, because the motion of the box has a constant velocity:

[tex]F-F_f-Wsin(\theta)=0\\\\[/tex]        (1)

F: force applied by the man = 95N

Ff: friction force

W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N

θ: degree of the inclined plane = 15°

You solve the equation (1) for Ff and you replace the values of all variables in the equation (1):

[tex]F_f=-Wsin(\theta)+F\\\\F_f=-(245N)sin18\°+95N=19.29N[/tex]

b) To fins the velocity of the box at the bottom you use the following formula:

[tex]W_N=\Delta K[/tex]   (2)

That is, the net work over the box is equal to the change in the kinetic energy of the box.

The net work is:

[tex]W_N=Mgsin(18\°)d-Ffd[/tex]

d: distance traveled by the box = 2.0m

[tex]W_N=245sin18\°(2.0m)N-19.29(2.0m)N=112.83J[/tex]

You use this value of the net work to find the final velocity of the box, by using the equation (2):

[tex]112.8J=\frac{1}{2}m[v^2-v_o^2]\\\\v_o=0m/s\\\\v=\sqrt{\frac{2(112.8J)}{m}}=\sqrt{\frac{225.67J}{25kg}}=3.00\frac{m}{s}[/tex]

The speed of the box, at the bottom of the incline plane is 3.00 m/s

Answer:

t = 25.5 min

Explanation:

To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.

[tex]t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h[/tex]

Next, you calculate the difference between both times t1 and t2:

[tex]\Delta t=t_1-t_2=2.30h-1.875h=0.425h[/tex]

This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:

[tex]0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s[/tex]

hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph

Answer:

It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4

Answer:

ma = 48.48kg

Explanation:

To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:

[tex]T=2\pi\sqrt{\frac{m_c}{k}}[/tex]     (1)

mc: mass of the chair

k: spring constant = 600N/m

T: period of oscillation of the chair = 0.9s

You solve the equation (1) for mc, and then you replace the values of the other parameters:

[tex]m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg[/tex]    (2)

Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:

T': period of chair when the astronaut is sitting = 2.0s

M: mass of the astronaut plus mass of the chair = ?

[tex]T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg[/tex] (3)

Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :

[tex]m_a=M-m_c=60.79kg-12.31kg=48.48kg[/tex]

The mass of the astronaut is 48.48 kg

Answer:

The minimum frequency of the coil is 7.1 Hz

Explanation:

Given;

number of turns, N = 200 turns

cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

magnitude of magnetic field strength, B = 30 x 10⁻³ T

maximum value of the induced emf, E = 8 V

Maximum induced emf is given as;

E = NBAω

where

ω is angular velocity (ω = 2πf)

E = NBA2πf

where;

f is the minimum frequency, measured in hertz (Hz)

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³  x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz

The minimum frequency of the coil in the case when it should be rotated in a uniform 30-mT magnetic field is 7.1 Hz.

Since

number of turns, N = 200 turns

cross-sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

the magnitude of magnetic field strength, B = 30 x 10⁻³ T

the maximum value of the induced emf, E = 8 V

Now

Maximum induced emf should be

E = NBAω

here,

ω is angular velocity (ω = 2πf)

Now

E = NBA2πf

here,

f is the minimum frequency

So,

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³  x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz.

Learn more about frequency here: https://brainly.com/question/24470698

From Ohm's law . . . Resistance = (voltage) / (current)

Resistance = (120 volts) / (7.6 Amperes)

Resistance = 15.8 Ω

Answer:

99.8

Explanation:

most massive the sun is at the center of the universe

Answer:

The correct option is D: "The small car and the truck experience the same average force."

Explanation:

The magnitude of the average force experienced by both bodies in motion is the same as explained by Newton's third law of motion. The force exerted by each body is equal and opposite in direction. The resulting acceleration experienced by each vehicle, however, will not be the same. It is greater for the small car.

Answer:

Vf = 78.64 m/s

Explanation:

The rocket is travelling upward at a constant acceleration of 3.99 m/s² until it runs out of fuel. So, in order to calculate its velocity at the point, where it runs out of fuel, we can simply use 3rd equation of motion:

2as = Vf² - Vi²

where,

a = acceleration = 3.99 m/s²

s = distance or height covered by rocket till fuel runs out = 775 m

Vf = Final Velocity = ?

Vi = Initial velocity = 0 m/s   (Since, rocket starts from rest)

Therefore,

2(3.99 m/s²)(775 m) = Vf² - (0 m/s)²

Vf = √(6184.5 m²/s²)

Vf = 78.64 m/s

Answer:

-2.24 m/s²

Explanation:

Given:

v₀ = 20 mi/hr = 8.94 m/s

v = 0 m/s

t = 4.0 s

Find: a

v = v₀ + at

0 m/s = 8.94 m/s + a (4.0 s)

a = -2.24 m/s²

Answer:

32mi

Explanation:

If 1lb contains 3,500 Cal

It means the number of hours required to burn 3500cal would be;

3500/331 = 10.57hours

But a brisk walk is 3.0 mi/h,

It means a distance of 3.0 × 10.57 mi would be covered = 31.71 miles

32miles{ approximated to the nearest whole}

Note Distance = speed × time

Answer:

[tex]\Delta p=1.3475\ kg-m/s[/tex]

Explanation:

The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-

We represent

The ball mass =  m = 275 g = 0.275 kg

Thus it goes to the floor and resurfaces upward.

The ball hits the ground at 3.30 m/s speed that is

u = -3.30 m/s which represents the Negative since the ball hits the ground)

It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)

[tex]\Delta p=p_f-p_i[/tex]

[tex]\Delta p=m(v-u)[/tex]

[tex]\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))[/tex]

[tex]\Delta p=1.3475\ kg-m/s[/tex]

Answer:

Approximately [tex]0.608[/tex] (assuming that [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex].)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

This answer will approach this question in two steps:

For simplicity, let [tex]a_{T}[/tex] represent the tangential acceleration ([tex]1.90\; \rm m \cdot s^{-2}[/tex]) of this car.

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to [tex]90^\circ[/tex] or [tex]\displaystyle \frac{\pi}{2}[/tex] radians.

The angular acceleration of this car can be found as [tex]\displaystyle \alpha = \frac{a_{T}}{r}[/tex]. ([tex]a_T[/tex] is the tangential acceleration of the car, and [tex]r[/tex] is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ([tex]u[/tex] and [tex]v[/tex]) to (tangential) acceleration [tex]a_{T}[/tex] and displacement [tex]x[/tex]:

[tex]v^2 - u^2 = 2\, a_{T}\cdot x[/tex].

The idea is to solve for the final angular velocity using the angular analogy of that equation:

[tex]\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta[/tex].

In this equation, [tex]\theta[/tex] represents angular displacement. For this motion in particular:

Solve this equation for [tex]\omega(\text{final})[/tex] in terms of [tex]a_T[/tex] and [tex]r[/tex]:

[tex]\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}[/tex].

Let [tex]m[/tex] represent the mass of this car. The centripetal force at this moment would be:

[tex]\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}[/tex].

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, [tex]m\, g[/tex].

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let [tex]\mu_s[/tex] denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

[tex]F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g[/tex].

The size of this force should be equal to that of the centripetal force when the car is about to skid:

[tex]\mu_s\, m\, g = \pi\, m\, a_{T}[/tex].

Solve this equation for [tex]\mu_s[/tex]:

[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g}[/tex].

Indeed, the expression for [tex]\mu_s[/tex] does not include any unknown letter. Let [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex]. Evaluate this expression for [tex]a_T = 1.90\;\rm m \cdot s^{-2}[/tex]:

[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608[/tex].

(Three significant figures.)

Answer: 10m

Explanation:

The magnitude of the vector would be 10

[tex]\sqrt{6^{2}+8^{2} } =10[/tex]

Answer:

4.09 MeV

Explanation:

Find the given attachment

Answer:

The difference between AC and DC lies in the direction in which the electrons flow. In DC, the electrons flow steadily in a single direction, or "forward." In AC, electrons keep switching directions, sometimes going "forward" and then going "backward."

Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

[tex]\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}[/tex]            (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

[tex]\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}[/tex]   (2)

dB/dt = -0.280T/s     (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

[tex]E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}[/tex]

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

Answer:

W = 414 J, correct is B

Explanation:

Work is defined by

        W = ∫ F .dx

where F is the force, x is the displacement and the point represents the dot product

this expression can also be written with the explicit scalar product

        W = ∫ F dx cos θ

where is the angle between force and displacement

for this case as the force is constant

         W = F x cos θ

calculate

         W = 25.0 18.0 cos (-23)

         W = 414 J

the correct answer is B